Functor to static member functions and type of static member function

Question

I was trying to assign a static member function to a functor as following:

 class myClass
 {
   public: 
     static void (myClass::*fptr)();
     static void myFun()
     {

      }
 };

 (void (myClass::*)()) myClass::fptr = &myClass::myFun; // doesn't work

 int main()
 {
   myClass obj;
   myClass::fptr = &myClass::myFun;   // neither does this works
   return 0; 
 }

The above assignment is not working.

At lest to check the type of the static member function(as I was not sure) I wrote the following statement in main

myClass::fptr = &myClass::myFun;

And the VS2010 intellisense displayed:

you can't assign a void ( * )() type to an entity of type void (myClass::*)()

I thought that type of static function should have been the latter(void ( * )())

I am also unable to figure out how to give definition for a static functor as it should be done for static data members.

I made the functor static because I want it to be independent of any instances of myClass.

Can somebody please elaborate on this topic of static functors and also on the type of the static member functions(as to why it is void(*)() in this case). Is it even possible to have a function pointer to a static member function?

EDIT:

I was able to work around this by using a typedef:

typedef void(*ptr)();
class myClass
{
 ....
 static ptr fptr;
};

ptr myClass::fptr = & myClass::myFunc;

But still why is the type static member function void( * )() ?

Jarod42 · Accepted Answer · 2013-08-29 15:11:14Z

0

As the method is static, you have to use a pointer on function instead of a pointer on method:

(void (*f_pointer)()) = &myClass::myFun;

answered Aug 29, 2013 at 15:11

Jarod42

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2 Comments

Uchia Itachi Over a year ago

Yes, I undestand. But why is the type the way it is?

Jarod42 Over a year ago

a (regular) method has an implicit argument (this). It is not the case for static method.

WormholeWizard · Accepted Answer · 2013-08-29 15:21:33Z

0

Class methods have one hidden argument - pointer to object of corresponding class, so you have to specify class in the pointer declaration. Static methods on other hand don't have such hidden argument, so you can work with them just like with regular functions.

This code will work:

class myClass
{
  public: 
    static void (*fptr)();
    static void myFun()
    {
      }
};

void (* (myClass::fptr))()  = &myClass::myFun;

answered Aug 29, 2013 at 15:21

WormholeWizard

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Comments

David Hammen · Accepted Answer · 2013-08-29 15:22:16Z

There are two key problems here.

The first is the declaration of myClass::fptr as static void (myClass::*fptr)(). This is a static data member that is a member function pointer that returns void. You want a static data member that is a function pointer that returns void. The declaration should be static void (*fptr)().

The other problem is something that is missing. You aren't defining (creating a space in memory for) that static data member. You need a file scope statement along the lines of void (*myClass::fptr)(); or void (*myClass::fptr)() = &myClass::myFun;

KasF · Accepted Answer · 2013-08-29 16:03:44Z

Static member functions behave like non-member functions. E.g. there is no this pointer in a static member function. Writing

void (MyClass::*func) () = ...

implifies the signature of the function looks like

void func(MyClass* thisObj);

But since there is no further parameter, you need to use ordinary syntax.

Use an ordinary function pointer:

void (*func)() = myClass::myFun;

Or better:

typedef void (*FPtr)();
FPtr func = myClass::myFun;

More better:

function<void()> func = myClass::myFun;

Better than more better:

auto func = myClass::myFun;

Edit: I just searched the standard but only found

(8.3.3/3) A pointer to member shall not point to a static member of a class (9.4), a 
member with reference type, or "cv void."

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